An organic compound 'A' C4H8 on treatment with KMnO4 /H+ yield's compound 'B' C3H6O. Compound 'A' al

Q: Why can't compound A be cyclobutane?

While cyclobutane also has the formula $C_4H_8$ (DU=1), it does not undergo ozonolysis or oxidative cleavage with $KMnO_4$. These reagents specifically target $\pi$-bonds in alkenes or alkynes.

Q: What is the difference between ozonolysis and $KMnO_4$ cleavage?

Reductive ozonolysis ($O_3$ followed by $Zn/H_2O_2$) stops at aldehydes and ketones. Acidic $KMnO_4$ is a stronger oxidant; it also cleaves the double bond but further oxidizes any formed aldehydes into carboxylic acids (and formaldehyde into $CO_2$ and $H_2O$).

Q: How did we know the other product of ozonolysis was formaldehyde?

Conservation of mass! Compound A has 4 carbons. Product B has 3 carbons. The remaining fragment must have $4 - 3 = 1$ carbon. The only 1-carbon carbonyl compound formed in ozonolysis is formaldehyde ($HCHO$).

Q: Why doesn't acetone get oxidized by $KMnO_4$?

Ketones lack a hydrogen atom attached to the carbonyl carbon, making them highly resistant to oxidation. Breaking a strong $C-C$ bond is required to oxidize a ketone, which doesn't happen under standard $KMnO_4$ cleavage conditions.

Q: What would be the product if compound A was but-2-ene?

But-2-ene ($CH_3-CH=CH-CH_3$) is a symmetrical alkene. Ozonolysis would yield two molecules of acetaldehyde ($CH_3CHO$). Acidic $KMnO_4$ would oxidize it to two molecules of acetic acid ($CH_3COOH$).

Question

An organic compound 'A' $C_4H_8$ on treatment with $KMnO_4 /H^+$ yield's compound 'B' $C_3H_6O$.
Compound 'A' also yields compound 'B' an ozonolysis.
Compound 'A' is

Options:

(A) 2-methylpropene

(B) 1-methylcyclopropane

(C) but-2-ene

(D) cyclobutane

Sigma Percentile · Accepted Answer

### Analyzing the Setup
The problem presents us with an unknown organic compound **A** having the molecular formula $C_4H_8$. The first thing we should always do when given a molecular formula is to calculate its **Degree of Unsaturation (DU)**.

The formula for DU is:
$$DU = C + 1 - \frac{H}{2} + \frac{N}{2} - \frac{X}{2}$$

Plugging in our values:
$$DU = 4 + 1 - \frac{8}{2} = 1$$

A degree of unsaturation of 1 implies that compound A could either be an alkene (containing one double bond) or a cycloalkane (containing one ring). However, the problem states that compound A undergoes **ozonolysis** and reacts with acidic $KMnO_4$. These are classic reactions of carbon-carbon double bonds, confirming that compound A is indeed an **alkene**.

### The Ozonolysis Clue
Compound A undergoes ozonolysis to yield compound **B**, which has the formula $C_3H_6O$.

Ozonolysis is a cleavage reaction that breaks the $C=C$ double bond, replacing it with two $C=O$ bonds. Since our starting alkene A has 4 carbon atoms and one of the products (B) has 3 carbon atoms, the other product must contain exactly 1 carbon atom. The only possible 1-carbon product from ozonolysis is **formaldehyde** ($HCHO$).

Now, what is compound B? With the formula $C_3H_6O$ and a DU of 1, it must be a carbonyl compound. The two possibilities are:
1. **Propanal** ($CH_3CH_2CHO$) - an aldehyde.
2. **Acetone** ($CH_3COCH_3$) - a ketone.

### The Master Equation: Acidic $KMnO_4$
To distinguish between propanal and acetone, we look at the second reaction. Compound A reacts with acidic $KMnO_4$ to yield the *exact same* compound B.

Acidic $KMnO_4$ is a powerful oxidizing agent. It cleaves the double bond similarly to ozonolysis but with a crucial difference: **it further oxidizes any resulting aldehydes into carboxylic acids**. Ketones, on the other hand, are resistant to further oxidation under these conditions and remain intact.

If compound B were propanal, the acidic $KMnO_4$ would have oxidized it into propanoic acid ($CH_3CH_2COOH$). Since compound B survives the strong oxidation, it **must be a ketone**. Therefore, compound B is **acetone** ($CH_3COCH_3$).

*(Note: The 1-carbon fragment, formaldehyde, is oxidized completely to $CO_2$ and $H_2O$ by $KMnO_4$.)*

### Final Calculation: Reconstructing Compound A
Now that we know the products of the cleavage are acetone and formaldehyde, we can easily reconstruct the original alkene A.

The trick is to place the oxygen atoms of the two carbonyl products face-to-face, remove them, and stitch the carbon atoms back together with a double bond:
$$(CH_3)_2C=O + O=CH_2 \rightarrow (CH_3)_2C=CH_2$$

The resulting structure is **2-methylpropene** (also known as isobutylene). This perfectly matches the molecular formula $C_4H_8$ and satisfies all the chemical reactions described in the problem.

Animated Solution for Chemistry - Hydrocarbons: An organic compound 'A' $C_{4} H_{8}$ on treatment with $K M n O_{4} / H^{+}$ yield's compound 'B' $C_{3} H_{6} O$ . Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is

Select Answer:

Visualized Solution

$C_{4} H_{8} \to C_{3} H_{6} O$

$D U = 1$

$C_{4} H_{8} O_{3} C_{3} H_{6} O + H C H O$

$Propanal or Acetone?$

$K M n O_{4} Oxidation$

$(C H_{3})_{2} C = O + O = C H_{2} \to (C H_{3})_{2} C = C H_{2}$

$2-methylpropene$

The Sigma Insight: Alkenes

Analyzing the Setup The problem presents us with an unknown organic compound A having the molecular formula $C_{4} H_{8}$

The Ozonolysis Clue

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C4​H8​→C3​H6​O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DU=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C4​H8​O3​​C3​H6​O+HCHO

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Propanal or Acetone?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KMnO4​ Oxidation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (CH3​)2​C=O+O=CH2​→(CH3​)2​C=CH2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2-methylpropene